Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.8 - Stokes'' Theorem - 16.8 Exercise - Page 1200: 17

Answer

$-32 \pi$

Work Step by Step

Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr $ The boundary of the surface is a circle with parameterization: $r=\lt 4 \cos t, 4 \sin t, 4 \gt$ and $dr = \lt -4 \sin t , 4 \cos t j, 0 \gt$ and $F(r(t))=\lt -4 \sin t , 4 \cos t j, -2k \gt$ Now, $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr =\int_{2 \pi}^{0} \lt -4 \sin t , 4 \cos t j, -2 \gt \cdot \lt -4 \sin t , 4 \cos t ,0 \gt dt\\=\int_{2 \pi}^0 16 \sin^2 t+16 \cos^2 t dt \\=16 \int_{2 \pi}^0 dt \\=-32 \pi$
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